The likelihoods of being dealt various poker hands are widely published (easily found on the internet). A more difficult problem is: what are the likelihoods of being dealt each poker hand, given a 54 card deck (52 card deck + 2 jokers).

The various hands of interest are:
1 pair
2 pair
3 of a kind
straight
flush
full house
4 of a kind
straight flush
5 of a kind

* Jokers can count as any rank card, in any suit.

(In reply to More results and solution by goFish)

I cant get the straights to count themselves, maybe you can point out where I've muddled up.

I approached this problem by considering three separate situations, one where you were dealt 2 wild cards, one where you were dealt 1 wild card, and one where you were dealt no wild cards.  I then pro-rated the calculated probabilities based on the respective probability of having been dealt that many wild cards.

The 2 blocks of text below are the brute force combinations that give a straight with one and two wild cards, suits not withstanding.  'X' is used to indicate a wild card, and the sets are ordered in decreasing order of natural cards.

A,K,Q,J,X
A,K,Q,X,T
A,K,X,J,T
A,X,Q,J,T
X,K,Q,J,T
K,Q,J,X,9
K,Q,X,T,9
K,X,J,T,9
X,Q,J,T,9
Q,J,T,X,8
Q,J,X,9,8
Q,X,T,9,8
X,J,T,9,8
J,T,9,X,7
J,T,X,8,7
J,X,9,8,7
X,T,9,8,7
T,9,8,X,6
T,9,X,7,6
T,X,8,7,6
X,9,8,7,6
9,8,7,X,5
9,8,X,6,5
9,X,7,6,5
X,8,7,6,5
8,7,6,X,4
8,7,X,5,4
8,X,6,5,4
X,7,6,5,4
7,6,5,X,3
7,6,X,4,3
7,X,5,4,3
X,6,5,4,3
6,5,4,X,2
6,5,X,3,2
6,X,4,3,2
X,5,4,3,2
5,4,3,X,A
5,4,X,2,A
5,X,3,2,A
X,4,3,2,A

A,K,Q,X,X
A,K,X,J,X
A,K,X,X,T
A,X,Q,J,X
A,X,Q,X,T
A,X,X,J,T
X,K,Q,J,X
X,K,Q,X,T
X,K,X,J,T
K,Q,X,X,9
K,X,J,X,9
K,X,X,T,9
X,X,Q,J,T
X,Q,J,X,9
X,Q,X,T,9
Q,J,X,X,8
Q,X,T,X,8
Q,X,X,9,8
X,X,J,T,9
X,J,T,X,8
X,J,X,9,8
J,T,X,X,7
J,X,9,X,7
J,X,X,8,7
X,X,T,9,8
X,T,9,X,7
X,T,X,8,7
T,9,X,X,6
T,X,8,X,6
T,X,X,7,6
X,X,9,8,7
X,9,8,X,6
X,9,X,7,6
9,8,X,X,5
9,X,7,X,5
9,X,X,6,5
X,X,8,7,6
X,8,7,X,5
X,8,X,6,5
8,7,X,X,4
8,X,6,X,4
8,X,X,5,4
X,X,7,6,5
X,7,6,X,4
X,7,X,5,4
7,6,X,X,3
7,X,5,X,3
7,X,X,4,3
X,X,6,5,4
X,6,5,X,3
X,6,X,4,3
6,5,X,X,2
6,X,4,X,2
6,X,X,3,2
X,X,5,4,3
X,5,4,X,2
X,5,X,3,2
5,4,X,X,A
5,X,3,X,A
5,X,X,2,A
X,X,4,3,2
X,4,3,X,A
X,4,X,2,A
X,X,3,2,A

These two sets count properly (42 with 1X and 63 with 2X) when determining how many straight flushes there are, but do not do so with regular straights, according to the pointed website statistics.

For clarity, there are 40 straight flushes possible with no wild cards, 332 (42*4suits*2wilds) with one wild card and 252 (63*4suits) with 2 wild cards, totalling 624, as expected.

With no wild cards, there are 10240 (10 possible sets *4suits *4suits * 4suits * 4suits * 4suits) possible straights, subtracting the 40 straight flushes gives 10200 ranked as a straight (which agrees with the conventional numbers).  This logic should carry forward to the one and two wild card cases, but I can't make it do so.

My math gives, for 1 wild card, 42*4*4*4*4*2 total straights (21504) with 1 wild and 63*4*4*4 total striaghts (4032) with 2 wilds.  Subtracting the 332+252 straight flushes and adding the  10200 from the no wild calculations gives 35152, not 34704 as indicated by the website and agreed by goFish, a difference of 448.

I've looked and looked and looked, but can't determine why I don't agree...

Posted by Cory Taylor on 2006-03-15 16:57:35